What is the resultant $p H$ $pH$ of the solution.....?

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What is the resultant $p H$ $pH$ of the solution.....?

What is the resultant $p H$ $pH$ of the solution prepared by taking a $10 \cdot m L$ $10mL$ volume of $H C l (a q)$ $HCl(aq)$ of $0.10 \cdot m o l \cdot L^{- 1}$ $0.10molL^-1$ concentration, and diluting this to a $1 \cdot L$ $1L$ volume?

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Answer 1 · 2017-04-21T08:08:38

$p H = 3$ $pH=3$

Explanation:

$p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_(10)[H_3O^+]$

$[H_{3} O^{+}] = \frac{10 \times 10^{- 3} \cdot L \times 0.1 \cdot m o l \cdot L^{- 1}}{1 \cdot L}$ $[H_3O^+]=(10xx10^-3*Lxx0.1*mol*L^-1)/(1*L)$

$10^{- 3} \cdot m o l \cdot L^{- 1}$ $10^-3*mol*L^-1$

And thus $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_(10)[H_3O^+]$

$-log_(10)(10^-3)=3...........$

And I did this all without a calculator. Note that normally, we would always add acid to water, and not water to acid. However, here the acid concentration is sufficiently dilute to get away with reverse addition.

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What is the resultant $p H$ $pH$ of the solution.....?